Exercise 3.1.2

• From equation (4), we have $E^{-1}=\left[\begin{array}{cc}\hfill I\hfill & \hfill -u{D}^{-1}\hfill \\ \hfill 0\hfill & \hfill D^{-1}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill I\hfill & \hfill 0\hfill \\ \hfill -v^T\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill I+u{D}^{-1}{v}^T\hfill & \hfill -u{D}^{-1}\hfill \\ \hfill -D^{-1}{v}^T\hfill & \hfill D^{-1}\hfill \end{array}\right]$
• From equation (5), we have

$E^{-1}=\left[\begin{array}{cc}\hfill {(I-u{v}^T)}^{-1}\hfill & \hfill 0\hfill \\ \hfill -v^T{(I-u{v}^T)}^{-1}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill I\hfill & \hfill -u\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {(I-u{v}^T)}^{-1}\hfill & \hfill -{(I-u{v}^T)}^{-1}u\hfill \\ \hfill -v^T{(I-u{v}^T)}^{-1}\hfill & \hfill 1+v^T{(I-u{v}^T)}^{-1}u\hfill \end{array}\right]$

We compare the 1,1 blocks to find $M^{-1}={(I-u{v}^T)}^{-1}$ in formula (1).